## Archive for the ‘Exercises’ Category

domestic arts in the age of digital distribution. RIP, vlorblog. here are some posts mentioning “clean” and “dish”: https://vlorblog.wordpress.com/?s=clean+dish (courtesy of the “search” feature from that blog).

our medium is handwriting. but

*don’t* trust even this little

bit of score; i can’t keep a beat

*even in real life*… not like a

drummer, say… with my rhythmic

body and *darn well* haven’t learned

how to put the *numbers* in.

syllable counts and beat counts just

blur up fast when i try to get a handle

on suchlike matters so i always give up

right away.

but scribbling out this display here

*was* helpful in learning to play the

damn thing out note-by-note. (i’ve been

wandering around the house singing this

for a few weeks now).

i got nothin

to say uh…

specially about

whatever was.

you can’t?

repeat the past?

of *course* y’can!

break on thoo

t’ th’ othuh side.

brae, brea,

brea, khah-nthoo.

bray, cahhn, thoo.

bray-con-thoo.

(to,

the other side).

is this some dagger

of the *mind*?

or…, well…,

what?

leaving only the “typing it up worse

for the net” to me… the -set

homework i’m about to loosely base my

plain-text solutions on is quite a lovely

thing to look upon… as my versions will

not be. now. make it so.

6.1.3

Let a/b, a’/b’, and a”/b” be consecutive fractions

in the Farey sequence of order n.

(i.e., in

F_n = {0/1, 1/n, …, 1/2, … , (n-1)/n, 1/1}

[for the record… it’s convenient to

have a “symbol” for this sequence;

F-subscript-n is standard but the

text doesn’t include it.]

)

We will show that these fractions

satisfy the “Mediant Property”:

a’/b’ = (a+a”)/(b+b”).

Consecutive “Farey fractions”

satisfy Theorem 6.1 and so

a’b – ab’ = 1 and

a”b’ – a’b” = 1.

Equating the left-hand sides,

a’b – ab’ = a”b’ – a’b”.

Regroup; factor;

“divide through by b'(b+b”)”:

a'(b+b”) = b'(a”+a)

a’/b’ = (a+a”)/(b+b”).

(We are done.)

in one sense, “textbook” exercises.

but in the usual sense, not.

i made up these “exercises” for myself,

but they sure as heck involve a *textbook*.

just now… very much in my usual way…

i made some pencil notes on a textbook page.

there was even a “warm-up”.

my *first* marks on that page…

“page 43”, let’s say… were to

scratch over the first “of” of

In number theory, the function of given in Theorem 2.2 would be called anand replace it with “ordered compositionofkwithnparts, because it gives a list ofnnumbers that add up tok.

*f*“.

because theorem 2.2, as it turns out,

tells us that

Theorem 2.2.The number of functionsffrom a set to the nonnegative integers such that isC(n + k – 1; k).

or, in other words, that

.

and that was, if you will, “exercise 1”:

“restate theorem 2.2 wordlessly”.

this code now appears in the (generous)

margin of page 43 (with the unimaginative

label “THEOREM 2.2”).

exercise 2?

well, the

bit appears *twice* on page 43

centered on a line of its own

(in a “display”, in other words).

so next to the first such display

i wrote out

(

*f* TAKES

*n* VALUES

ADDING UP

TO *k*

)

… where the parens “(“,”)” indicate

big ol’ *handwritten* parens

bracketing the whole “block”

of handwritten “f takes n

values adding up to k” thing;

it fills the “display” space

(to the right of the displayed code)

rather nicely from top to bottom.

so the first “exercise” was of the form:

“translate from english into code”

and the second was of the form

“translate from code into english”.

“write it up in ” will here

be considered *beyond the scope of our
post*… it *is* time for breakfast…

i just couldn’t wait to observe here that such

“translating”, back-and-forth, code-to-language,

language-to-code is a *large part* of “doing math”.

throw in code-to-graph (and back) and get even more.

venn-diagrams back-and-forth to truth tables.

(hey, look. “isomorphisms”!)

anyhow. we’re *always* doing suchlike exercises.

trouble is. what-you-mean-“we”?

*feed* me!

6.

With a TI “grapher”, I’ve just found that

244^2 == 5 (mod 1009). One simply puts

Y_1 = 1009(fPart(X^2/1009))…

this computes the equivalence-class-mod-1009

of the square of each input successively…

and “scrolls down” in the TABLE window

(set to display natural number X’s).

Hence x^2 == 5 (mod 1009) *does* have a solution.

(In fact, two of them; the other is of course

765 (= 1009 – 244)).

In detail,

244^2 = 58536 = 59*1009 + 5

and

765^2 = 585225 =580*1009 + 5.

(The point here is that one may easily

*verify* such results by “paper-&-pencil”

methods.)

To find the roots *entirely* by p-&-p methods,

the best line of attack would probably be via

“primitive roots”.

*trouble is, is, that that “5”
was supposed to’ve been “150”.
once more, dear friends!
(or close the wall up with our
english dead!)*

ladies and gentlemen, PSL(2,7).

Let p be an odd prime.

Let L = (1^2)(3^2)…([p-1]^2) and

let R = (-1)^{(p+1}/2}.

We will show that L = R (mod p).

Note that for 1 \le k < p

("\le" denotes is-less-than-or-equal-to)

we have k = -(p-k) (mod p).

Rearranging the factors of L gives

L=1(-1)(p-1)2(-1)(p-2)…(p-2)(-1)(2).

L=(-1)^{(p-1)/1}1*2*3*…*(p-1).

Since 1*2*…*(p-1) = (p-1)!, we can

apply Wilson's Theorem:

L\equiv (-1)^{(p-1)/2}*(-1) (mod p).

But then L = R as we were to show.

One may now derive

(2^2)(4^2)…([p-1]^2) \equiv R

by squaring "Wilson's Equation"

[(p-1)!]^2 \equiv (-1)^2 (mod p).

Note that (-1)^2 = 1 = (-1)^{p-1}:

[(p-1)!]^2 \equiv(-1)^{p-1} (mod p).

Divide on the left by L;

divide on the right by R.

The result follows.

(Division "works" because p

is a prime… we can always

"cancel" factors less than p

in this case.)

(One should *not*… if instructed

to prove *two* equivalences…

simply remark that the second

"is similar to" the first:

if our textbook authors hadn't

wanted at least *some* details,

they presumably would not have

asked that part of the question

at all.)

(n!+1, (n+1)!+1)=1. [For all n \in N.]

[Let n \in N.]

Suppose (n!+1, (n+1)!+1) = K > 1.

Then K has a prime divisor p>1.

We have p|(n!+1) and p|(n+1)!+1.

Since “p|x and p|y” implies that

“p|(y-x)” in all cases, we have

p|[ (n+1)!+1 – (n!+1) ]

p|[ (n+1)n! – n!]

p|n*n!.

But then (because p is prime) p|n!.

This is absurd since we also know p|n!+1.

No such p exists; K = 1; we are done.

[

i’ve been cranking out *lots* of number

theory problems… i’ve never taught the

subject at this level of detail and *need*

to do lots of problems (in order that

exercises like this can *become* “routine”.)

but i’ve only *typed up* a few (one

problem set [out of four that i’ve marked

so far]). anyhow, none of the class

had a proof this short-&-sweet so here

it is now for my loyal subscribers.

this is a *great* quarter so far.

getting back to work.

oops, “semester” so far.

\bye

]